# Consider two insulated chambers (A,B) of same volume connected by a closed knob, S. One mole of perfect gas is confined in chamber A. What is the change in entropy of gas when knob S is opened? ( R= 8.31 )

## I can't understand for which process the entropy has to be calculated. The question says insulated chamber which indicates that it is an adiabatic process but entropy is constant for an adiabatic process. Please explain what to do. Thanks.

Jul 17, 2018

An isentropic ($\Delta S = 0$) process is both

• adiabatic (ie no transfer of heat or matter between system and surroundings ... so this is adiabatic)

and

• reversible $\textcolor{red}{\boldsymbol{\leftarrow}}$ the key idea here

Your intuition should tell you that the gas will spontaneously flow through the opened knob and occupy both chambers A & B. That spontaneity is synonymous with irreversibility, and the arrow of time.

To illustrate, as the system contains 1 mole of perfect gas, the probability of every particle being in chamber $A$ (as opposed to chamber $B$, or vice-versa ) at any reasonable time after the knob is opened is:

• ${\left(\frac{1}{2}\right)}^{{N}_{A}} \approx {10}^{- {10}^{23.3}}$

It is possible but, in practical terms, impossible - without outside interference.

The system can be returned to its original state... using a reversible process ... that involves work being done on the gas, and heat being released from the gas to its surroundings to keep the temperature constant, quasi-statically, we can even use the thermodynamic definition of entropy.

But the quicker way to explore change in entropy is to split the volume $V$ of chamber $A$ into micro-volumes, each of $\mathrm{dV}$, which can only accommodate one particle of gas, so that:

• $N = \frac{V}{\mathrm{dV}}$

The number $\Omega$ of possible micro-states in this somewhat discretised system is:

• Omega = ""^NP_n qquad "where "{(n = "no of particles of gas"), (""^NP_n = (N!)/((N-n)!) ):}

Doubling the available volume means:

$N ' = \frac{2 V}{\mathrm{dV}} = 2 N$

:. Omega' = ""^(2N)P_n

From Boltzman's definition of entropy:

• $S = {k}_{B} \ln \Omega$

The change in entropy is:

$\Delta S = {k}_{B} \left(\ln \Omega ' - \ln \Omega\right)$

= k_b(ln(( ""^(2N)P_n)/( ""^(N)P_n))) ~~ k_B ln ((2N )/N)^n qquad "using Stirling approximation"

$= n {k}_{B} \ln 2$

For one mole: $n = {N}_{A}$, and ${N}_{A} {k}_{B} = R$

So :

• $\Delta S = R \ln 2$

You can construct a reversible process that takes the system from its initial to final state, or vice versa, but that process is no longer adiabatic.

Final thought:

• $\left\{\begin{matrix}\mathrm{dS} = \frac{{\mathrm{dQ}}_{\textcolor{red}{r e v}}}{T} \\ \mathrm{dS} \ge \frac{\mathrm{dQ}}{T}\end{matrix}\right.$